Problem 3 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-14: Solve the equation by factoring. $$ 15 x^{2}-12=-8 x $$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{2}{3}\) and \(x = -\frac{6}{5}\).

Step 1: Move Terms to One Side

Start by moving the term on the right side of the equation to the left side. We have the equation: \[ 15x^2 - 12 = -8x \] Add \(8x\) to both sides:\[ 15x^2 + 8x - 12 = 0 \] Now the equation is set to zero.

Step 2: Identify and Factor the Trinomial

We need to factor the quadratic expression \(15x^2 + 8x - 12\). We look for two numbers that multiply to \(15 \times (-12) = -180\) and add to \(8\). These numbers are \(18\) and \(-10\), since \(18 \times (-10) = -180\) and \(18 + (-10) = 8\).

Step 3: Rewrite and Group Terms

Rewrite the middle term \(8x\) using the numbers found: \[ 15x^2 + 18x - 10x - 12 = 0 \] Now, group the terms:\[ (15x^2 + 18x) + (-10x - 12) = 0 \]

Step 4: Factor by Grouping

Factor out the greatest common factor from each group:\[ 3x(5x + 6) - 2(5x + 6) = 0 \] Notice that \(5x + 6\) is a common factor now. This gives:\[ (3x - 2)(5x + 6) = 0 \]

Step 5: Solve Each Factor for Zero

Set each factor equal to zero and solve for \(x\): \(3x - 2 = 0\):\[ 3x = 2 \]\[ x = \frac{2}{3} \] \(5x + 6 = 0\):\[ 5x = -6 \]\[ x = -\frac{6}{5} \] The solutions are \(x = \frac{2}{3}\) and \(x = -\frac{6}{5}\).

Key Concepts

Factoring

Factoring is a vital algebraic skill used to solve quadratic equations, such as the one given in the exercise. When we factor, we are essentially breaking down an algebraic expression into its simpler components, known as factors, which when multiplied together give the original expression. This is akin to finding the pieces that fit together to recreate a puzzle.

For the quadratic equation in the exercise, the expression we need to factor is \(15x^2 + 8x - 12\). The goal in factoring this expression is to find two binomials that multiply to reconstruct the original quadratic expression.

• First, identify two numbers that multiply to \(15 \times (-12) = -180\) and add up to \(8\). These numbers are \(18\) and \(-10\).
• Rewrite the middle term \(8x\) using these numbers: \(18x - 10x\).
• Group terms into pairs and factor each group:\((15x^2 + 18x), (-10x - 12)\).
• Factor out the greatest common factors: \(3x(5x + 6) - 2(5x + 6)\).
• Notice the common binomial factor \(5x + 6\): \((3x - 2)(5x + 6)\).

This process of factoring is crucial because it transforms a complex expression into simpler ones, allowing us to solve the equation more easily.

Solving Equations

Solving equations involves finding the variable values that make the equation true. In the case of quadratic equations like the one you've encountered here, solving requires setting the factored expression equal to zero and solving each resulting linear equation. Let's break down this process.

After factoring the quadratic equation to \((3x - 2)(5x + 6) = 0\), the next step is to use the Zero Product Property. This property states if a product of two factors is zero, at least one of the factors must also be zero.

• Set each factor equal to zero:
• For \(3x - 2 = 0\), add \(2\) to both sides to get \(3x = 2\), then divide by \(3\) to find \(x = \frac{2}{3}\).
• For \(5x + 6 = 0\), subtract \(6\) from both sides to obtain \(5x = -6\), then divide by \(5\) to discover \(x = -\frac{6}{5}\).

By solving these linear equations, you determine the values of \(x\) that satisfy the original quadratic equation. Hence, the solutions are \(x = \frac{2}{3}\) and \(x = -\frac{6}{5}\), showing you all possible values for the variable.

Algebraic Expressions

Algebraic expressions form the backbone of algebra and consist of constants, variables, and arithmetic operations combined. The quadratic equation from the exercise is a particular type of algebraic expression. Breaking it down helps to understand its function and structure.

The expression \(15x^2 + 8x - 12\) is called a trinomial because it consists of three terms:

• \(15x^2\) is the quadratic term, where \(15\) is the coefficient and \(x^2\) indicates the degree (second degree).
• \(8x\) is the linear term, where \(8\) is the coefficient and \(x\) is the variable.
• \(-12\) is the constant term, representing a fixed value unaffected by \(x\).

Understanding how to manipulate these expressions is essential in solving not just quadratic equations but broader algebra problems too. The capabilities developed through algebraic manipulation play a key role in finding solutions and simplifying even complex expressions.